As the title already says it I have this expression $$ f(s)=\sum_{n=1}^\infty \frac{n^{s-1}}{e^{2\pi n}-1} $$ and am wondering if this one can be expressed in terms of any special or analytical functions?

Note that for large $n$ the denominator behaves like $e^{2\pi n}$ and thus approximately the main contributions come from the terms around the maximum $$n=\frac{s-1}{2\pi} $$ so that it should behave like $\sim (s-1)^{s-1}$.

In fact asymptotically it seems like $$(2\pi)^{s}f(s) \sim \Gamma(s) \, .$$

Thank you

**Update:** In writing $f(-s+1)$ I re-expressed $$n^{-s}= \frac{1}{\Gamma(s)}\int_{0}^\infty t^{s-1} \, e^{-nt} \, {\rm d}t \, ,$$ so that
\begin{align}
f(-s+1) &= \sum_{n=1}^\infty \frac{n^{-s}}{e^{2\pi n} -1} \\
&= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}t \, t^{s-1} \sum_{n=1}^\infty e^{-nt} \sum_{k=0}^\infty e^{-2\pi n(k+1)} \\
&= \frac{1}{\Gamma(s)} \int_0^\infty {\rm d}t \, t^{s-1} \sum_{k=0}^\infty \frac{1}{e^{t} \, e^{2\pi (k+1)} -1} \\
\end{align}
and maybe $$\sum_{k=1}^\infty \frac{1}{z \, e^{2\pi k}-1}$$ is simpler to evaluate?

More than the precise values of $f(s)$ I'm actually interested in the analytic structure. Is it possible to see whether there are any zeros?

**2nd Update:** Being made aware about the Mellin-Transform Method shown at various sites here on MSE by Marko Riedel I carried out the analysis for certain positive $s$ values. For negative $s$ some more information can be found here: A new formula for Apery's constant and other zeta(s)?

We adapt the notation from above at little and write $$ f_k(x) = \sum_{n=1}^\infty \frac{n^{k-1}}{e^{nx}-1} $$ with Mellin-Transform $$ {\cal M}_{f_k}(s) = \Gamma(s)\zeta(s)\zeta(1+s-k) $$ where the integral converges for $s>k$. The inverse Mellin is given by $$ f_k(x)=\frac{1}{2\pi i} \int_{k+\frac{1}{2}-i\infty}^{k+\frac{1}{2}+i\infty} \Gamma(s)\zeta(s)\zeta(1+s-k) \, x^{-s} \, {\rm d}s \, . $$ We now close the contour at ${\rm Re}(s)=-\frac{1}{2}$ and transform this integral. First $$ \int_{-\frac{1}{2}+i\infty}^{-\frac{1}{2}-i\infty} \Gamma(s)\zeta(s)\zeta(1+s-k) \, x^{-s} \, {\rm d}s \\ \stackrel{u=k-s}{=} -\int_{k+\frac{1}{2}-i\infty}^{k+\frac{1}{2}+i\infty} \Gamma(k-u)\zeta(k-u)\zeta(1-u) \, x^{u-k} \, {\rm d}u $$ and then using the $\zeta$-functional equation $$ \zeta(1-s) = \frac{2}{\left(2\pi\right)^s} \, \cos\left(\frac{\pi s}{2}\right) \Gamma(s) \zeta(s) $$ and continue (renaming $u$ back to $s$) $$ =(-1)^{\frac{k}{2}+1}\int_{k+\frac{1}{2}-i\infty}^{k+\frac{1}{2}+i\infty} \Gamma(s)\zeta(s)\zeta(1+s-k) \, \left(2\pi\right)^{-s} \, \left(\frac{x}{2\pi}\right)^{s-k} \, {\rm d}s $$ where we needed to assume $k$ even. Obviously the value $x=2\pi$ is somewhat special and plugging in this becomes $$ =(-1)^{\frac{k}{2}+1} \, 2\pi i \, f_k\left(2\pi\right) \, . $$

The poles we enclosed are at $s=k$, $s=1$ and $s=0$ and the residue theorem yields $$ \left(1-\left(-1\right)^{\frac{k}{2}}\right) f_k(2\pi) = \frac{1}{2\pi i} \oint_\gamma \Gamma(s)\zeta(s) \zeta(1+s-k) \, (2\pi)^{-s} \, {\rm d}s \\ = \frac{\Gamma(k)\zeta(k)}{(2\pi)^k} + \frac{\Gamma(1)\zeta(2-k)}{2\pi} + \zeta(0)\zeta(1-k) \, . $$ In this equation not all even $k$ values are allowed, but only those of the form $$ k=4n+2 \qquad n=0,1,2,3,... $$ and we obtain the result $$ f_{4n+2}(2\pi) = \frac{(4n+2)!\zeta(4n+2)}{2(4n+2)(2\pi)^{4n+2}} + \frac{\zeta(-4n)}{4\pi} - \frac{\zeta(-4n-1)}{4} \\ = \frac{{\rm B}_{4n+2}}{2(4n+2)} - \frac{\delta_{n,0}}{8\pi} \, . $$

In fact the latter result also seems to be very close for the numbers $$ k=4n+4 \qquad n=0,1,2,3,... $$ if we correct the negative sign of these Bernoulli numbers $$ f_{2n}(2\pi) = \frac{(-1)^{n+1} \, {\rm B}_{2n}}{2(2n)} - \frac{\delta_{n,1}}{8\pi} \qquad n=1,2,3,... \, . $$